Skydiver's velocity with retarding forces

1. The problem statement, all variables and given/known data
An 80 kg skydiver jumps from an altitude of 1000m and opens his chute at 200m
(a) The total retarding force on the diver is a constant 50.0N without his chute open, it is constant 3600N with his chute open. What will be the skydiver's speed when he reaches the ground?

3. The attempt at a solution
Well, frankly I've gone through several methods and it's all become a big mess. It seems like a straight forward problem but I think I've gotten tunnel vision from it and am missing something obvious. I have a hunch I should be using PE and KE rather than the following methods, but this was all I could come up with.

I don't think you need PE or KE. You know the force that is slowing him down and it is constant so just find the acceleration due to this force, and subtract it from the acceleration due to gravity.
This will be your overall acceleration now, and just use projectile motion equations. Then once he passes 200m you need to find the new overall acceleration because the force changes.

But keep in mind that when he opens his parachute he will have some initial velocity.

1. The problem statement, all variables and given/known data
An 80 kg skydiver jumps from an altitude of 1000m and opens his chute at 200m
(a) The total retarding force on the diver is a constant 50.0N without his chute open, it is constant 3600N with his chute open. What will be the skydiver's speed when he reaches the ground?

3. The attempt at a solution
Well, frankly I've gone through several methods and it's all become a big mess. It seems like a straight forward problem but I think I've gotten tunnel vision from it and am missing something obvious. I have a hunch I should be using PE and KE rather than the following methods, but this was all I could come up with.

Here's how I would do it
from 1000m to 200m
upwards force=50N
downwards force=W=mg=784.8N
so net force is 734.8N downwards.
F=ma
solve to get a (from 1000m to 200m), which I found to be -9.185m/s^2
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, for which I got 121.2m/s (at 200m) However, if you are supposed totake into account approximate terminal velocity of a human, roughly 54m/s, you might simply need to replace 121.2m/s with that.
So now we can use the final velocity of that part as initial velocity in the next part
From 200m to 0m
upwards force = 3600N
downwards force=W=mg=784.8N
so net force = 2815.2N upwards
F=ma
solve for a (from 200m to 0m), which I found to be 35.19m/s^2 (upwards)
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, as we did previously for the first length, and there's your answer (~2.2m/s if you took into account my approximate value of human terminal velocity)

Here's how I would do it
from 1000m to 200m
upwards force=50N
downwards force=W=mg=784.8N
so net force is 734.8N downwards.
F=ma
solve to get a (from 1000m to 200m), which I found to be -9.185m/s^2
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, for which I got 121.2m/s (at 200m) However, if you are supposed totake into account approximate terminal velocity of a human, roughly 54m/s, you might simply need to replace 121.2m/s with that.
So now we can use the final velocity of that part as initial velocity in the next part
From 200m to 0m
upwards force = 3600N
downwards force=W=mg=784.8N
so net force = 2815.2N upwards
F=ma
solve for a (from 200m to 0m), which I found to be 35.19m/s^2 (upwards)
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, as we did previously for the first length, and there's your answer (~2.2m/s if you took into account my approximate value of human terminal velocity)

Presuming Terminal veloicity is beyond the scope of the problem.

Besides you don't need any simplifying assumptions. If you bother to run the numbers you arrive at the same answer, that you can derive more simply.